Absorption of a simple random walker A 1D random walk with absorbing barriers at 0 and N. Can we determine the mean time before an absorption event takes place at either barriers given that the probability to move left/right is $0.5$ everywhere except at barriers, where then the probability to be absorbed in next jump is $p$ and the probability to jump away from the barrier is $q?$  where $p>q$.
In the simpler case where p and q are also equal to 0.5, the mean time is known, $t_a=a(N-a)$ where $a$ is the starting site of the walker and 0 and N are the positions of the two barriers.
 A: Many interesting points have already been pointed out in the comments, amongst others by user Did, which in essence already contain an answer to the post, so I decided to write them up here as an answer for future readers.

Given that the absorbing barriers are placed at $0$ and $N,$ we have the following important transition probabilities: $N-1\to N-2$ happens with probability $q$ and $N-1\to N$ with probability $p$ ($q+p=1$), similarly, the transition $1\to 2$ happens with probability $q$ and $1\to 0$ with probability $p.$ All remaining transitions happen with probability $1/2.$ Thus we are only modifying the behaviour at the barriers, and asking how favouring absorption influences the mean time before an absorption event takes place. 
Now we perform a Markov decomposition after one step of the walk, this yields the mean time before absorption $t_1 = 1+qt_2$ starting the walk from $1,$ $t_{N-1}=1+qt_{N-2}$ starting from $N-1$ and for every other starting point $2\leqslant k \leqslant N-2 ,$ the corresponding mean time is $t_k = 1+\frac{1}{2}(t_{k-1}+t_{k+1}).$ The latter condition implies that $t_k$ ought to take the form of 
$$t_k=uk+v-k^2\tag{1}$$ 
for some couple $(u,v)$ independent of $k.$ Moreover, the former conditions on $t_1$ and $t_{N-1}$ imply the following relations: 
$$
\begin{align}
u+v-1 =& 1+q(2u+v-4) \tag{2} \\
u(N-1)+v-(N-1)^2 =& 1+q(u(N-2)+v-(N-2)^2). \tag{3}
\end{align}
$$
The equations $(1), (2)$ and $(3)$ together form a system of equation that can be solved, yielding: $$u=N\qquad v=((q/p)-1)(N-2)$$ hence the modified mean time before absorption is given by: 
$$
t_a = a(N-a) + ((q/p)-1)(N-2)
$$
for every starting point $a$ such that $1\leqslant a\leqslant N-1.$
